Moving point problems in junior high school mathematics As shown in the figure, the side length of diamond ABCD is 6cm, and the angle B is 60 °. From the beginning, point P and Q start from point a at the same time. Point P moves in the direction of a to C to B at the speed of 1cm / s. point Q moves in the direction of a to B to C to D at the speed of 2cm / s. when point Q moves to point d, P and Q stop moving at the same time, The time for Q to move is x seconds, and the area of the overlapping part of △ Apq and △ ABC is y square centimeter (it is stipulated here that the point and line segment are triangles with an area of 0). The following questions are answered: (1) the time taken for point P and Q to start and meet is & nbsp; & nbsp; & nbsp; seconds. (2) the process of point P and Q from starting to stopping. When △ Apq is an equilateral triangle, the value of X is & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &Find the functional relationship between Y and X

Moving point problems in junior high school mathematics As shown in the figure, the side length of diamond ABCD is 6cm, and the angle B is 60 °. From the beginning, point P and Q start from point a at the same time. Point P moves in the direction of a to C to B at the speed of 1cm / s. point Q moves in the direction of a to B to C to D at the speed of 2cm / s. when point Q moves to point d, P and Q stop moving at the same time, The time for Q to move is x seconds, and the area of the overlapping part of △ Apq and △ ABC is y square centimeter (it is stipulated here that the point and line segment are triangles with an area of 0). The following questions are answered: (1) the time taken for point P and Q to start and meet is & nbsp; & nbsp; & nbsp; seconds. (2) the process of point P and Q from starting to stopping. When △ Apq is an equilateral triangle, the value of X is & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &Find the functional relationship between Y and X

Based on the velocity and trajectory of points P and Q, the following analysis can be carried out: connect the BD intersection AC to point O, AC = 6cm. Before P moves to point O, it is obvious that PA = 1cm / s * x (s) = x (CM) (the unit of X is omitted in the following calculation, s, the unit of Y is square centimeter, and the length of line segment is centimeter) Pb = 2 * x = 2x. It is known that in △ PAQ, ∠ PAQ = 60 degrees

How many ways to solve the problem of moving point in junior high school mathematics?

The moving point problem of junior high school mathematics is generally related to the area of the figure and the judgment of the figure
It can be combined with equation, function and inequality
And can be divided into "single moving point" and "two moving points", not a few sentences can solve
I suggest you ask a specific question

The problem of moving point in junior high school mathematics
Who has a math problem? Help me

[05] as shown in the figure, in the right angle trapezoid ABCD, ad ‖ BC, ∠ C = 90 °, BC = 16, DC = 12, ad = 21. The moving point P starts from point D and moves at a speed of 2 units per second along the direction of ray da. The moving point Q starts from point C and moves to point B at a speed of 1 unit per second on the line CB. The points P and Q start from point D and C at the same time, Then point P stops moving. Let the time of motion be t (seconds)
(1) Let the area of △ bpq be s, and find the functional relationship between S and t;
(2) What is the value of T when a triangle with B, P and Q vertices is isosceles triangle?
(3) When line segment PQ and line segment AB intersect at point O and 2ao = ob, the tangent value of ∠ BQP is obtained;
(4) Is there a time t such that PQ ⊥ BD? If so, find out the value of T; if not, explain the reason
[solution] (1) as shown in Figure 3, if the crossing point P is PM ⊥ BC and the perpendicular foot is m, then the quadrilateral pdcm is a rectangle
∵QB＝=6-t,∴S=(1/2)×12×(16-t)=96-t
(2) It can be seen from the figure that: cm = PD = 2T, CQ = T. the triangle with B, P and Q as the vertex is isosceles triangle, which can be divided into three cases:
① If PQ = BQ. In RT △ PMQ, pq2 = T2 + 122, T2 + 122 = (16-t) 2 is obtained from pq2 = bq2, and T = 7 / 2 is obtained;
② If BP = BQ, in RT △ PMB, bp2 = (16-t) 2 + 122
(16-2t) 2 + 122 = (16-t) 2, that is 3t2-32t + 144 = 0
Because Δ = - 704

As shown in the figure, the quadrilateral ABCD is rectangular, ad = 16cm, ab = 6cm. The moving points P and Q start from a and C at the same time. Point P moves to D at the speed of 3cm / s until D, and Q moves to B at the speed of 2cm / S. (1) a few seconds after P and Q start from the start, the area of quadrilateral abqp is 35 times of the rectangular area product? When is the area of quadrilateral abqp the largest and what is the largest? (2) A few seconds after P and Q start, PQ = 65cm?

(1) The area of rectangle ABCD is s = 16 × 6 = 96cm2, 35S rectangle = 35 × 96 = 57.6cm2. After x seconds, the area of quadrilateral abqp is 35 times that of rectangle, that is 12 (3x + 16-2x) × 6 = 35 × 96, and the solution is x = 3.2 seconds

Is there a simple algorithm for (1997 + 1996 * 1998) / (1997 * 1998-1)?

1997*1998-1=（1996+1）*1998-1=1996*1998+1*1998-1=1996*1998+1997,
The denominator is equal to the numerator, so the answer is 1

On the volume of a rotating body of higher number
The volume of a graph composed of x ^ 2 + y ^ 2 ≤ 2x and Y ≥ x to circle x = 2. One of the options is 2 π ∫ {0,1} (2-x) [(2x-x ^ 2) ^ 1 / 2-x]. I think this is right, but the answer is not this. Is this correct

The speed of a car increases by 10 MGS from the beginning to the first 100 m distance, and the speed increases by ()
A. 4.1m/sB. 8.2m/sC. 16.4m/sD. 32.8m/s

The acceleration of uniform acceleration linear motion: a = v122x = 1002 × 100M / S2 = 0.5m/s2. According to: V22 − V12 = 2aX: V2 = V12 + 2aX = 100 + 2 × 0.5 × 100M / S = 102m / s. then the increase of velocity: △ v = 102 − 10 ≈ 4.1m/s. So a is correct, B, C and D are wrong. So a

Simple calculation of 6.3 × 99

6.3*99
=6.3*(100-1)
=6.3*100-6.3*1
=630-6.3
=623.7

Given that x.y.z satisfies the equation system x + 2y-4z = 0 2x + y-5z = 0, we can find the value of X: Y: Z
Thank you

x+2y-4z=0 ①
2x+y-5z=0 ②
① * 2 - ②：y = z ③
③ Substituting ①: x = 2Z
x:y:z=2:1:1

A 48 meter long iron wire is surrounded into a rectangle, and its length is three times of its width. What is the area of the rectangle?

Length + width = 48 △ 2 = 24m
So width = 24 △ 3 + 1 = 6m
Length = 24-6 = 18 m
Area = 6x18 = 108 square meters = 10800 square decimeters